\(\Gamma \)  has infinite diameter since a single surgery can change the rank of the first homology at most by one.
To show that the outdegree of \(\Gamma \)  is infinite, let \(v_{(M,\xi )}\)  be a vertex of \(\Gamma \) . For an even integer \(n\) , choose a Legendrian knot \(L\)  in a Darboux-ball in \((M,\xi )\)  with Thurston–Bennequin invariant \(\operatorname{tb}=1\)  and rotation number \(\operatorname{rot}=n\) . Denote the contact manifold obtained by contact \((-1)\) -surgery along \(L\)  by \(L(-1)\) . A calculation as for example in [1]}, [2]} shows that the homology of \(L(-1)\)  is
\(H_1\big (L(-1)\big )=H_1(M)\oplus \mathbb {Z}_{\mu _L},\) 
where the \(\mathbb {Z}\) -summand is generated by a meridian \(\mu _L\)  of \(L\)  and that the Poincaré dual of the Euler class \(\operatorname{e}(L(-1))\)  is given by
\(\operatorname{e}(L(-1))=\operatorname{e}(M,\xi )+n\mu _L,\) 
where \(\operatorname{e}(M,\xi )\)  denotes the Poincaré dual of the Euler class of \((M,\xi )\) . Thus, we get infinitely many different contact manifolds by contact \((-1)\) -surgery from \(M\) . The same construction with contact \((+1)\) -surgery along a Legendrian knot with Thurston–Bennequin invariant \(\operatorname{tb}= -1\)  provides the fact that the indegree is also infinite.
